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        <h1 id="考研数二"><a href="#考研数二" class="headerlink" title="考研数二"></a>考研数二</h1><h2 id="高数资料篇"><a href="#高数资料篇" class="headerlink" title="高数资料篇"></a>高数资料篇</h2><span id="more"></span>
<h3 id="公式"><a href="#公式" class="headerlink" title="公式"></a>公式</h3><h4 id="泰勒公式"><a href="#泰勒公式" class="headerlink" title="泰勒公式"></a>泰勒公式</h4><script type="math/tex; mode=display">
\begin{align} 
f(x)    &=\sum_{n=0}(n!)^{-1}f^{(n)}(x_0)(x-x_0)^n\\
sinx     &=x-\frac {x^2}{3!}+o(x^2) \\
cosx     &=1-\frac {x^2}{2!}+\frac {x^4}{4!}+o(x^4) \\
tanx     &=x+\frac {x^3}{3}+o(x^3)\\
arcsinx &=x+\frac {x^3}{3!}+o(x^3)\\
arctanx &=x-\frac {x^3}{3}+0(x^3)\\
e^x     &=1+x+\frac {x}{2!}+\frac {x}{3!}+o(x^3)\\
ln(1+x) &=x-\frac {x^2}{2}+\frac {x^3}{3}+o(x^3)\\
(1+x)^a &=1+ax+\frac {a(a-1)}{2!}x^2+0{x^2}\\
\end{align}</script><h3 id="题型解法"><a href="#题型解法" class="headerlink" title="题型解法"></a>题型解法</h3><h4 id="解微分方程"><a href="#解微分方程" class="headerlink" title="解微分方程"></a>解微分方程</h4><h5 id="1-一阶微分方程"><a href="#1-一阶微分方程" class="headerlink" title="1.    一阶微分方程"></a>1.    一阶微分方程</h5><h6 id="类型一-可分离变量型"><a href="#类型一-可分离变量型" class="headerlink" title="类型一 可分离变量型"></a>类型一 可分离变量型</h6><ol>
<li>​     $y^{‘}=f(x)·g(y)$</li>
</ol>
<p>2.</p>
<script type="math/tex; mode=display">
y^{'}=f(ax+by+c) \\
令\ u=ax+by+c\Rightarrow u^{'}=a+bf(u)\\
\int \frac {du}{a+bf(u)}=\int dx</script><h6 id="类型二-一阶线性"><a href="#类型二-一阶线性" class="headerlink" title="类型二 一阶线性"></a>类型二 一阶线性</h6><p>能写成  $y^{‘}+p(x)y=q(x)$  则为<strong>一阶线性微分方程</strong>(详解请看P288)</p>
<script type="math/tex; mode=display">
通解为:\quad
y=e^{-\int{p(x)dx}}\left[\int {e^{-\int{p(x)dx}}\cdot q(x)dx}+C\right]</script><h5 id="2-二阶可微分方程"><a href="#2-二阶可微分方程" class="headerlink" title="2.    二阶可微分方程"></a>2.    二阶可微分方程</h5><h6 id="类型一-二阶可降阶微分方程"><a href="#类型一-二阶可降阶微分方程" class="headerlink" title="类型一 二阶可降阶微分方程"></a>类型一 二阶可降阶微分方程</h6><p>能写成  $y^{‘’}=f(x,y^{‘})$  的形式<em>(三步)</em></p>
<script type="math/tex; mode=display">
\begin{aligned}
&第一步:令\ y^{'}=p,\ y^{''}=p^{'}\Rightarrow 原方程\ \frac {dp}{dx}=f(x,p)\\
&第二步:求得通解\ p=y^{'}= \psi(x,C_1)  \\
&第三步:用一阶微分方程的解法\\
\end{aligned}</script><p>能写成 $y^{‘’}=f(y,y^{‘})$ 的形式<em>(三步)</em></p>
<script type="math/tex; mode=display">
\begin{aligned}
&第一步 令y^{'}=p,则\ y^{''}=p^{'}=\frac{dp}{dy}\cdot\frac{dy}{dx}=\frac{dp}{dy}\cdot p,则原方程就变为关于\ y\ 的一阶方程\frac{dp}{dy}\cdot p=f(y,p)\\
&第二部使用一阶方程的方法得到通解\ p=y^{'}= \phi (x,C_1) ,又将\ p\ 还原得到关于\ x\ 的一阶方程\\
&第三步同理得出通解
\end{aligned}</script><h6 id="类型二-二阶常系数微分方程"><a href="#类型二-二阶常系数微分方程" class="headerlink" title="类型二 二阶常系数微分方程"></a>类型二 二阶常系数微分方程</h6><p>能写成   $y^{‘’}+qy^{‘}+py=f(x)$  叫做<strong>高阶常系数微分方程</strong>(两步)</p>
<script type="math/tex; mode=display">
\begin{equation}
\begin{aligned}
第一步\ \lambda^2+p\lambda+q=0得出\lambda_1\ ,\lambda_2\ 写出其次方程的通解:\\
\end{aligned}
\end{equation}</script><script type="math/tex; mode=display">
\begin{cases}
若\ p^2-4q>0\ 通解:\ y=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}\\
若\ p^2-4q=0\ 通解:\ y=(C_1+C_2x)e^{\lambda x}\\
若\ p^2-4q<0\ 通解:\ y=e^{ax}(C_1cos\beta x+C_2sin\beta x)\quad 根为a\pm\beta i\\
\end{cases}</script><script type="math/tex; mode=display">
\begin{aligned}
第二步 设特解y^{\ast}\Rightarrow 得待定系数\Rightarrow 得特解\Rightarrow 写通解\\
\end{aligned}</script><script type="math/tex; mode=display">
\quad\begin{cases}
    &当f(x)=P_n(x)e^{ax},\\
    &设y^{\ast} =e^{ax}Q_n(x)x^k\\
    &其中k=\begin{cases}
            0,\quad a\neq\lambda_1\neq\lambda_2,\\
            1,\quad a=\lambda_1 或a=\lambda_2,\\
            2,\quad a=\lambda_1 =\lambda_2,\\
        \end{cases}
\end{cases}</script><script type="math/tex; mode=display">
\begin{cases}
    &当f(x)=e^{ax}\left[P_m(x)cos\beta x+P_n(x)sin\beta x\right]时\\
    &设y^{\ast} =e^{ax}\left[Q_l^{(1)}cos\beta x+Q_l^{(2)}sin\beta x\right]x^k\ ;\\
    &\quad\quad\quad\quad其中k=\begin{cases}
            0,a\pm\beta 不是特征根,\\
            1,a\pm\beta 是特征根,\\
        \end{cases}
\end{cases}</script><h5 id="3-“-y-n-n-geq3-”情况"><a href="#3-“-y-n-n-geq3-”情况" class="headerlink" title="3.    “$y^{(n)},n\geq3$”情况"></a>3.    “$y^{(n)},n\geq3$”情况</h5><p>p_296</p>
<h3 id="归纳总结"><a href="#归纳总结" class="headerlink" title="归纳总结"></a>归纳总结</h3><h4 id="中值定理"><a href="#中值定理" class="headerlink" title="中值定理"></a>中值定理</h4><ol>
<li><p>设f<strong>(x)在[a,b]上连续</strong>，则</p>
<script type="math/tex; mode=display">
 \begin{aligned}
     &定理1（有界与最值定理）m≤f(x)≤M,其中，m,M分别为f(x)在[a,b]上的最小值与最大值。\\
     &定理2（介值定理）当m≤\mu≤M时，存在\xi∈[a,b]们，使得f（\xi）=\mu\\
     &定理3（平均值定理）当a<x_1<x_2<\cdot\cdot\cdot<x_n<b时，在[x_1,x_n]内至少存在一点\xi，使得\\
     &\quad\quad\quad\quad\quad\quad\quad f(\xi)=\frac {f(x_1)+f(x_2)+\cdot\cdot\cdot+f(x_n)}{n}\\
     &定理4（零点定理）当f(a)·f(b)<0时，存在\xi∈(a,b),使得f(\xi)=0.
 \end{aligned}</script></li>
<li><p>涉及导数（微分）的中值定理</p>
<script type="math/tex; mode=display">
\begin{aligned}
&定理5（费马定理）设f(x)满足在x_0点处\left\{\begin{aligned}&1.可导;\\&2.取极值;\end{aligned}\right.\ 则f^{'}(x_0)=0\ ;\\
&定理6（罗尔定理）设f(x)满足\left\{\begin{aligned}&1.在[a.b]上连续;\\&2.在(a,b)上导;\\&3.f(a)=f(b);\\&注:3替换为\lim_{x \rightarrow a^+}f(x) = \lim_{x \rightarrow b^-}f(x)也成立;\end{aligned}\right.\ 则存在\xi\in(a,b),使得f^{'}(\xi)=0\ ;\\
&定理7（拉格朗日中值定理）\ 设f(x)满足\left\{\begin{aligned}&1.在[a.b]上连续;\\&2.在(a,b)内可导;\end{aligned}\right.\ 则存在\xi\in(a,b),使得f(b)-f(a)=f^{'}(\xi)(b-a)\ ;\\
&定理8（柯西中值定理）\ 设f(x)满足\left\{\begin{aligned}&1.在[a.b]上连续;\\&2.在(a,b)内可导;\\ &3.g^{'}(x)\neq0;\end{aligned}\right.\ 则存在\xi\in(a,b),使得\frac {f(b)-f(a)}{g(b)-g(a)}=\frac{f^{'}(\xi)}{g^{'}(\xi)}\ ;
\end{aligned}</script></li>
</ol>
<p><strong>关系:</strong></p>
<script type="math/tex; mode=display">
柯西中值定理\stackrel{当g(x)=x时}\Rightarrow 拉格朗日中值定理 \stackrel{加入罗尔的条件三}\Rightarrow 罗尔中值定理\\
罗尔中值定理\Rightarrow 拉格朗日中值定理 \Rightarrow 柯西中值定理\\
.....\\
三个定理都可以互相论证,由此可以看出是一回事,证明时循环论证.</script><h4 id="反常积分"><a href="#反常积分" class="headerlink" title="反常积分"></a>反常积分</h4>
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